Question

If P(X) = x/6, what are the possible values of X for it to be a probability distribution?
A. 0, 2, 3
B. 1, 1, 2
C. 2, 3, 4
D. 1, 2, 3

PLEASE ANSWER MY QUESTION PROPERLY. I REALLY NEED A DECENT ANSWER

Answer 1

Step-by-step explanation:

What does a probability mean?

It means that :-

$\rm{\dfrac{ Number \ of \ Possible \ Outcomes}{Total \ Outcomes}}$

Now, going parts by parts:-

A) part is not applicable as zero cannot be a possible answer.

B) part, 2 same occurences cannot take place back to back. It's wrong.

C) Let us check this as C and D are possible. But, we need to check it!

$\sf{\dfrac{2 + 3 + 4 }{6}}$

$\sf{\dfrac{9}{6}}$ is not a possible answer.

D) part, let's check:-

$\sf{\dfrac{1 + 2 + 3 }{6}}$

$\sf{\dfrac{6}{6}}$ =1, So, D) part is the correct answer.

Answer 2

Answer:

Given :-

• P(X) = x/6

Find Out :-

• What are the possible values of X for it to be a probability distribution.

Solution :-

➪ We know that,

$\tt{Possibility(A) =\:\dfrac{Number\: of\: Favourable\: Outcome}{Total\: Number\: of\: Favourable\: Outcomes}}$

Now, let's check the given options which one is correct :

◘ Option No A : 0 , 2 , 3

$\Rightarrow$ The part does not apply because zero can't be a correct answer.

So, Option no A is incorrect.

◘ Option No B : 1 , 1 , 2

$\Rightarrow$ The same events can't place together.

So, Option No B is also incorrect.

◘ Option No C : 2 , 3 , 4

$\Rightarrow$$\tt{\dfrac{x}{6}}$

$\Rightarrow$$\tt{\dfrac{2 + 3 + 4}{6}}$

$\Rightarrow$$\tt{\dfrac{5 + 4}{6}}$

$\Rightarrow$$\tt{\dfrac{9}{6}}$

So, Option No C is also incorrect.

◘ Option No D : 1 , 2 , 3

$\Rightarrow$$\tt{\dfrac{x}{6}}$

$\Rightarrow$$\tt{\dfrac{1 + 2 + 3}{6}}$

$\Rightarrow$$\tt{\dfrac{3 + 3}{6}}$

$\Rightarrow$$\tt{\dfrac{6}{6}}$

$\Rightarrow$ $\small\purple{\underline{{\boxed{\textbf{1}}}}}$

Hence, the answer we get 1.

So, Option No D is correct option.